Theorem: Let $w\colon\mathbb{R}^2\rightarrow\mathbb{R}$ a function that satisfies the characteristic equation $A(w_x)^2+Bw_x w_y+C(w_y)^2=0$ associated to the PDE with constant coefficients
$$A\frac{\partial{^2u}}{\partial{x}^2}+B\frac{\partial{^2u}}{\partial{x}\partial{y}}+C\frac{\partial{^2u}}{\partial{y^2}}+D\frac{\partial{u}}{\partial{y}}+E\frac{\partial{u}}{\partial{x}}+Fu=G \tag 1$$
Let $k$ constant. Prove if $w(x,y)=k$ is a characteristic curve of $(1)$ if and only if $w(x,y)=k$ is a solution of the equation
$A\left(\frac{dy}{dx}\right)^2-B\frac{dy}{dx}+C=0 \tag 2$
I found the proof in a book but i don't understand some parts of the proof.
Proof:
$(\implies )$
Suppose $w_y\not = 0$.
By hypothesis $w(x,y)=k$ is a characteristic curve of $(1)$ then $w$ satisfies the equation $$A(w_x)^2+Bw_xw_y+C(w_y)^2=0 \tag 3$$
By implicit theorem we have:
$$\frac{dy}{dx}=-\frac{w_x}{w_y} \tag 4$$
Dividing each term of $(3)$ by $(w_y)^2$ we have:
$$A\left(\frac{w_x}{w_y}\right)^2+B\frac{w_x}{w_y}+C=0 \tag 5$$
Then, by $(4)$ we have:
$$A\left(\frac{dy}{dx}\right)^2-B\frac{dy}{dx}+C=0 $$
In this proof. I cannot understand how he use the theorem of implicit function. I don't see that the hypotheses are fulfilled in order to use the theorem.
($\Leftarrow$)
Suppose $w_y\not = 0$
Then, by implicit function theorem we have:
$\frac{dy}{dx}=-\frac{w_x}{w_y} \tag 6$
By hypothesis we have $w(x,y)=k$ is a solution of the equation $(2)$
Then, by $(6)$ we have:
$0=A(\frac{dy}{dx})^2-B\frac{dy}{dx}+C=A(\frac{w_x}{w_y})^2+B\frac{w_x}{w_y}+C$
Then
$A(\frac{w_x}{w_y})^2+B\frac{w_x}{w_y}+C=0$
This conclude the proof.
In this last step i'm a very confused. Why is this enough?
$A(\frac{w_x}{w_y})^2+B\frac{w_x}{w_y}+C=0$
The other question is how use the implicit function theorem.
Thanks for all!